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Given two algebraic varieties $X$ and $Y$, the Künneth theorem implies that there is a relation between $H^*(X) \otimes H^*(Y)$ and $H^*(X \times Y)$, and in fact in many cases they are equal.

Given an open subvariety $U \subseteq X \times Y$ and a closed subvariety $Z \subseteq U$, is there a way to extend this and properly decompose $H^{i+j}_Z(U)$ into the parts that are "morally" from $H^i(X) \otimes H^j(Y)$?

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    $\begingroup$ Say that $X$ and $Y$ are affine spaces. With no hypotheses on $U$ or $Z$, $H^\ast_Z(U)$ can be arbitrarily complicated and I don't think it makes sense to talk abour parts that 'come from' the (trivial) cohomologies of $X$ and $Y$. $\endgroup$ Commented Jun 9, 2017 at 21:57
  • $\begingroup$ Well, I'm thinking of etale cohomology anyways... $\endgroup$ Commented Jun 9, 2017 at 22:03
  • $\begingroup$ What I said is valid in étale cohomology, too. $\endgroup$ Commented Jun 9, 2017 at 22:14

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In the topology case, it is the Kunneth theorem for excision pair.In the algebraic geometry(etale cohomology) case , it can be reduced to prove that for $Z_1$,$Z_2$(if $Z=Z_1\times Z_2)$and sheaf $Ri_1^! F$and $Ri_2^!G$ since $R\Gamma(Z,Ri^!F)=R\Gamma(X,R\Gamma_ZF)$.Kunneth theorem needs the projection formula and base change theorems.If we only consider the case that the base scheme is spec k,the generic base change theorem in SGA4.5 is always available。$Ri^!F$ should be locally constant sheaf due to the projection formula.If Z and X are smooth,this is from purity theorems. The case that X is smooth and Z is a divsor can be found in SGA7 XIII lemma 2.1.10.

In fact the topology purity theorem is often proved by using Kunneth theorem for excision pair(c.f Milnor's characteristic class


update: I had thought you just consider the Künneth case.To give the relationship between the cohomology for general closed subscheme Z and the $pr_{1*}Z$ $pr_{2*}Z$ is related to Motives and algebraic correspondences.In the case that $Z=Z_1\times Z_2$,Künneth formula can expressed in the form $H(Z)=Hom(H(Z_1),H(Z_2))$ by Poincare duality (transfer tensor to hom),When Z is not as this form and $H_C(X,A)$ corresponds to$c^!RHom(p_1^*A_2,p_2^!A_2)=RHom(c_1^*A_1,c_2^!A_2)$ by kunneth theorem (c.f SGA5 III), in the case that Z corresponds to the graph of a morphism,it is isomorphic to$RHom(f^*A_1,A_2)$,we can calculate the $H_C(X,A)$ by Fubini-like theroem in SGA5 III proposition 3.4.But I don't really know how wide can this theorem apply

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  • $\begingroup$ Sounds like it's time to dig up SGA... $\endgroup$ Commented Jun 9, 2017 at 22:12
  • $\begingroup$ Actually, wait, does this only apply in the case $Z=Z_1 \times Z_2$? $\endgroup$ Commented Jun 12, 2017 at 19:01
  • $\begingroup$ @W. Schlieper update $\endgroup$
    – wongdl
    Commented Jun 13, 2017 at 6:15
  • $\begingroup$ @W.Schlieper Yes this answer is only about the case $Z=Z_1 \times Z_2$. As I wrote in my comment a few days ago there is nothing you can say at the level of generality that you asked the question. $\endgroup$ Commented Jun 13, 2017 at 8:56

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